This is the first of a series of posts on elementary Lie theory. My goal is to reach the classification of simple Lie algebras by Dynkin diagrams starting from scratch. I will first recall the various basic definitions.

Definition:Lie algebra is a vector space $\mathfrak{g}$ over some field $\mathbb{K}$ together with a bilinear application $\mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}$ $(x,y) \mapsto [x,y]$

satisfying the two following axioms:

1. $[x,x] = 0$ for all $x,y,z \in \mathfrak{g},$
2. $[[x,y],z] + [[y,z],x] + [[z,x],y] = 0$ for all $x,y,z \in \mathfrak{g}.$

The second axiom is called the Jacobi identity. Since this application is not necessarily associative, it is essential to include the Lie bracket when writing the product. Note that bilinearity of the product and the first axiom give us $0 = [x+y,x+y] = [x,x] + [x,y] + [y,x] + [y,y]$ $= [x,y] + [y,x]$

for all $x,y \in \mathfrak{g}.$ So multiplication in a Lie algebra is anticommutative. This means that for all $x,y \in \mathfrak{g},$ $[x,y] = -[y,x].$

Here is an important

Example: Starting with any associative algebra $A$ over $\mathbb{K}$ (i.e. a $\mathbb{K}$-vector space with an associative bilinear multiplication), we can create a Lie algebra $[A]$ by defining $[x,y] := xy - yx.$ Then $[x,x] = 0$ and $[ax_1 + bx_2,y] = (ax_1 + bx_2)y - y(ax_1 + bx_2)$ $= ax_1y + bx_2y - yax_1 - ybx_2$ $= a(x_1y - yx_1) + b(x_2y - yx_2)$ $= a[x_1,y] + b[x_2,y]$

for all $a,b \in \mathbb{K}$ and $x_1, x_2, y \in A.$ By the same argument on the other variable, we get that $(x,y) \mapsto [x,y]$ is bilinear as hoped. Note that we only used bilinearity of multiplication in $A$ to show this. However, the associativity condition will be seen to be necessary (and sufficient) for the Jacobi identity to be valid. Indeed, for any $x,y,z \in A,$ we have $[[x,y],z] = [(xy - yx),z]$ $= (xy - yx)z - z(xy - yx)$ $= (xy)z - (yx)z - z(xy) + z(yx)$

thus $[[x,y],z] + [[y,z],x] + [[z,x],y] = (xy)z - (yx)z - (yx)z + z(yx)$ $+ (yz)x - (zy)x - x(yz) + x(zy)$ $+ (zx)y - (xz)y - y(zx) + y(xz)$ $= (xy)z - x(yz) + y(xz) - (yx)z$ $+ (zx)y - z(xy) + z(yx) - (zy)x$ $+ (yz)x - y(zx) + x(zy) - (xz)y.$

Therefore, the Jacobi identity is satisfied if and only if multiplication in $A$ is associative.

Definition: For any algebra $\mathfrak{g},$ there is a linear map $\text{ad} : \mathfrak{g} \to \text{End}_{\mathbb{K}} \mathfrak{g}$ defined by $\text{ad}(x)(y) = [x,y].$

In fact, $\text{ad}(x)$ is nothing else than left multiplication by $x$ in $\mathfrak{g}.$

Definition: derivation of an algebra $A$ is a linear map $D : A \to A$ satisfying the Leibniz rule : $D(xy) = D(x)y - xD(y).$

Proposition: Let $A$ be an algebra over $\mathbb{K}$ satisfying axiom 1 ( $[x,x] = 0$ for all $x \in A$ with $[ , ]$ denoting multiplication in $A$). Then $A$ is a Lie algebra iff $\text{ad}(x)$ is a derivation of $A$ for all $x \in A.$

Proof: Suppose $[x,x] = 0$ in $A.$ Then using the anticommutativity of $A,$ the Jacobi identity reads $[z,[x,y]] = [[y,z],x] + [[z,x],y]$ $= -[x,[y,z]] + [[z,x],y]$ $= [x,[z,y]] + [[z,x],y]$

which is equivalent to $\text{ad}(z)([x,y]) = [x,\text{ad}(z)(y)] + [\text{ad}(z)(x),y].$

QED

We have showed that a Lie algebra is an anticommutative algebra in which left multiplication by any element is a derivation.

Let’s now state some more

Definitions: A Lie algebra homomorphism is a linear map $\varphi : \mathfrak{g} \to \mathfrak{h}$ between two Lie algebras such that $\varphi([x,y]) = [\varphi(x), \varphi(y)]$ and a Lie algebra isomorphism is a bijective homomorphism. If $\mathfrak{a}, \mathfrak{b}$ are subsets of $\mathfrak{g}$, define $[\mathfrak{a}, \mathfrak{b}] := \text{span} \{[x,y] : x \in \mathfrak{a}, y \in \mathfrak{b} \}.$

So a Lie subalgebra $\mathfrak{h}$ of $\mathfrak{g}$ is a subspace such that $[\mathfrak{h},\mathfrak{h}] \subset \mathfrak{h}$, i.e. it is asked to be closed under multiplication in $\mathfrak{g}$. An ideal of $\mathfrak{g}$ is a subspace $\mathfrak{h}$ such that $[\mathfrak{h},\mathfrak{g}] \subset \mathfrak{h}$.

Note that for two subspaces $\mathfrak{a}$ and $\mathfrak{b}$, the anticommutativity of the bracket gives that $[\mathfrak{a},\mathfrak{b}] = [\mathfrak{b},\mathfrak{a}]$ (because for $x \in \mathfrak{a}$ and $y \in \mathfrak{b}$, we have that $[x,y] = -[y,x] = [-y,x] \in [\mathfrak{b},\mathfrak{a}])$ so taking products of subspaces is commutative and there is no distinction between left and right ideals.

Finally, with $\mathfrak{g}$ a Lie algebra and $\mathfrak{a}$ an ideal of $\mathfrak{g},$ we can form the quotient Lie algebra $\mathfrak{g}/\mathfrak{a}$ by defining $[x+\mathfrak{a}, y + \mathfrak{a}] = [x,y] + \mathfrak{a}.$

References:

1. Lectures on Lie Groups and Lie algebras by R. Carter, G. Segal, I. MacDonald.
2. Lie Groups Beyond an Introduction by W. Knapp.